If x^{m} y^{n}=(x+y)^{m+n},then frac{dy}{dx} | vedclass

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(D) Given that,$x^{m} y^{n}=(x+y)^{m+n}$.Taking the natural logarithm on both sides,we get:$m \ln x + n \ln y = (m+n) \ln (x+y)$Differentiating both s

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$\frac{y}{x}$

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(D) Given that,$x^{m} y^{n}=(x+y)^{m+n}$.Taking the natural logarithm on both sides,we get:$m \ln x + n \ln y = (m+n) \ln (x+y)$Differentiating both sides with respect to $x$,we have:$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = (m+n) \frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} + \frac{m+n}{x+y} \frac{dy}{dx}$$\left(\frac{n}{y} - \frac{m+n}{x+y}\right) \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x}$$\left(\frac{nx + ny - my - ny}{y(x+y)}\right) \frac{dy}{dx} = \frac{mx + nx - mx - my}{x(x+y)}$$\left(\frac{nx - my}{y(x+y)}\right) \frac{dy}{dx} = \frac{nx - my}{x(x+y)}$Canceling the common term $(nx - my)$ from both sides:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x}$Therefore,$\frac{dy}{dx} = \frac{y}{x}$.

If $x^{m} y^{n}=(x+y)^{m+n}$,then $\frac{dy}{dx}$ is equal to

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